Evaluate: $$\int_{-\infty}^{\infty}\frac{\cos x}{1+x^6}$$
I have found many examples on similar questions with -- $1+x^2$ -- in the denominator but I am not sure how the higher power would affect the solution. Any help or solution would be much appreciated
On
The function $f(z)={e^{iz}\over{1+z^6}}$ has 6 singularities of order 1 and residue $1\over {6z_0}$ (consider $lim_{z\to z_0}(z-z_0)f(z)$) where $z_0=e^{i({\pi\over 6}+{\pi\over 3}k)}$. If you define $C$ as a semicircle on half positive-$im$ of $\Bbb C$ of radius $R$ i.e. define $C$ as follows: $$C=C_1\cup C_2$$ $$C_1=\lbrace{z=x|-R\le x\le R}\rbrace$$ $$C_2=\lbrace{z=Re^{i\theta}|0\le\theta\le\pi}\rbrace$$ then you can write: $$\int_C f(z)dz=\int_{C_1} f(z)dz+\int_{C_2} f(z)dz=\int_{-R}^{R} f(x)dx+\int_{0}^{\pi} {Rie^{i\theta}{{e^{iRe^{i\theta}}}\over{1+R^6e^{6i\theta}}}}d\theta={{2\pi}\over{3}}$$
by using residue theorem. Tending $R\to\infty$ makes the second integral zero and we by taking real from both sides we get: $$I={{2\pi}\over{3}}$$
You proceed as in the case with $1+x^2$ in the denominator. Define $$ f(z)=\frac{e^{iz}}{1+z^6} $$ and integrate along a semicircle of radius $R$ on the upper half-plane, use the residue theorem, and let $R\to\infty$. The poles go $f$ are the solutions of the equation $1+z^6=0$.